Consider a bounded sequence whose terms are eventually "outgrown" by other terms of the sequence, i.e., for any index there exists an such that . Does this property imply that is eventually monotonic, i.e., that there exists an index such that is monotonically increasing from index onwards? I can't think of any counterexample. If true, do you know if this a known theorem which I can reference from some standard textbook?

NB: the boundedness assumption is essential, otherwise you can construct simple counterexamples like

Thanks,
jens

May 31st 2012, 06:46 AM

girdav

Re: Eventually self-outgrowing = eventually monotonic?

and .

May 31st 2012, 03:49 PM

jens

Re: Eventually self-outgrowing = eventually monotonic?

Thanks for your counterexample!
Sorry for the mess, but I had only sequences in mind whose variations become "tiny", that is, whose and coincide. But I forgot to tell. So let me correct my question:
Assume converges to a finite limit and for any index , there exists such that . Does this imply that there exists an index such that for any two and both larger than , we have ?
Now I hope this statement is correct (and non-trivial!)